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adelina 88 [10]
2 years ago
8

3. Two people push on a car with a force of 5150 N, and the car moves a distance of 8.0m. What is the

Physics
2 answers:
MArishka [77]2 years ago
6 0
I think c not sure tho
tangare [24]2 years ago
3 0
The answer is D, work done is equal to force times it’s distance travelled, so 5150*8.0=41200J
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Find the acceleration a body whose velocity increases from 11m/s to 33m/s in 10 seconds
solong [7]

Answer:

1.1 m/(s)^2

Explanation:

u=11 m/s

v=33 m/s

t=10s

v=u+at

=> 33=22+(a)(10)

=> 33-22=10a

=> 10a=11

=> a=11/10=1.1 m/(s)^2

7 0
2 years ago
George walks to a friend's house. He walks 850 meters north before he realizes he walked too far.
anyanavicka [17]

Answer:

<h2>44 m/s</h2>

Explanation:

In this problem we are expected to calculate the velocity of Georges movements.

Given data

Total distance covered by  George= 850+250= 1100 meters

Time taken  by  George to cover the total distance= 25 seconds

We know that velocity is, v= distance/ time

Therefore substituting our data into the expression for velocity we have

v= 1100/ 25= 44 m/s

Hence the velocity in m/s is 44

7 0
2 years ago
A hammer strikes one end of a thick iron rail of length 8.80 m. A microphone located at the opposite end of the rail detects two
stepladder [879]

Answer:

ΔT = 0.02412 s

Explanation:

We will simply calculate the time for both the waves to travel through rail distance.

FOR THE TRAVELING THROUGH RAIL:

T_{rail} = \frac{Distance}{Speed\ of\ Sound\ in\ Rail}\\\\T_{rail} = \frac{8.8\ m}{5950\ m/s}\\\\T_{rail} = 0.00148\ s

FOR THE WAVE TRAVELING THROUGH AIR:

T_{air} = \frac{Distance}{Speed\ of\ Sound\ in\ Air}\\\\T_{air} = \frac{8.8\ m}{343\ m/s}\\\\T_{air} = 0.0256\ s

The separation in time between two pulses can now be given as follows:

\Delta T = T_{air}-T_{rail} \\\Delta T = 0.0256\ s - 0.00148\ s\\

<u>ΔT = 0.02412 s</u>

3 0
2 years ago
How do I answer this question
maria [59]

Answer:

what question

Explanation:

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